Page 22.3: Dragomir-Type Integral Inequalities

Theorem (Dragomir Integral Inequality)

Let \(x, f, g\in \mathcal{C}(J)\) be nonnegative functions, \(L\in \mathcal{C}(J\times \mathbb{R}_+)\) be a nonnegative function such that
\begin{equation*}
0\leq L(t, x)-L(t, y)\leq M(t, y)(x-y)
\end{equation*}
for any $t\in J$ and for any  $x\geq y\geq 0$, where $M\in \mathcal{C}(J\times\mathbb{R})$ be a nonnegative function. If
\begin{equation*}
x(t)\leq f(t)+g(t) \int_a^t L(s, x(s))\Delta s,\quad t\in J,
\end{equation*}
then
\begin{equation*}
x(t)\leq f(t)+g(t) \int_a^t h_1(s) e_{\ominus h_2}(\sigma(s), t) \Delta s,\quad t\in J,
\end{equation*}
where
\[\begin{eqnarray*}
h_1(t)&=& L(t, f(t)),\\ \\
h_2(t)&=& g(t)M(t, f(t)),\quad t\in J.
\end{eqnarray*}\notag\]

Proof
Let
\begin{equation*}
z(t)= \int_a^t L(s, x(s))\Delta s,\quad t\in J.
\end{equation*}
Then
\begin{equation}
\label{5.2} x(t)_\leq f(t)+g(t) z(t),\quad t\in J,
\end{equation}
and
\begin{equation*}
z(0)=0.
\end{equation*}
We have
\[\begin{eqnarray*}
z^{\Delta}(t)&=& L(t, x(t))\\ \\
&\leq& L(t, f(t)+g(t) z(t))\\ \\
&=& L(t, f(t)+g(t)z(t))-L(t, f(t))+L(t, f(t))\\ \\
&\leq& M(t, f(t))\left(f(t)+g(t) z(t)-f(t)\right) +L(t, f(t))\\ \\
&=& L(t, f(t))+g(t) M(t, f(t)) z(t)\\ \\
&\leq& h_1(t)+h_2(t) z(t),\quad t\in J.
\end{eqnarray*}\notag\]
Hence, we get
\begin{equation*}
z(t)\leq \int_a^t h_1(s) e_{\ominus h_2}(\sigma(s), t)\Delta s,\quad t\in J.
\end{equation*}
From here and from \eqref{5.2}, we obtain
\begin{equation*}
x(t)\leq f(t) +g(t) \int_a^t h_1(s) e_{\ominus h_2}(\sigma(s), t) \Delta s,\quad t\in J.
\end{equation*}
This completes the proof.

Theorem (Dragomir Integral Inequality)

Let \(x, f, g\in \mathcal{C}(J)\) be nonnegative functions, \(L\in \mathcal{C}(J\times\mathbb{R}_+)=\) be a nonnegative function, \(\psi\in \mathcal{C}(\mathbb{R}_+)\)  be a nonnegative strictly increasing function with \(\psi(0)=0\) and
\begin{equation*}
0\leq L(t, x)-L(t, y)\leq M(t, y) \psi^{-1}(x-y)
\end{equation*}
for any \(t\in J\) and for any \(x\geq y\geq 0\), where \(M\in \mathcal{C}(J\times\mathbb{R}_+)\) be a nonnegative   function and \(\psi^{-1}\) is the inverse of \(\psi\). If
\begin{equation*}
x(t)\leq f(t)+\psi\left(g(t)\int_a^t L(s, x(s))\Delta s\right),\quad t\in J,
\end{equation*}
then
\begin{equation*}
x(t)\leq f(t)+\psi\left(g(t)\int_a^t h_1(s) e_{\ominus h_2}(\sigma(s), t)\Delta s\right),\quad t\in J,
\end{equation*}
where \(h_1\) and \(h_2\)  are defined as above.

Proof
Let
\begin{equation*}
z(t)=\int_a^t L(s, x(s)) \Delta s,\quad t\in J.
\end{equation*}
Then
\begin{equation}
\label{5.5} x(t)\leq f(t)+\psi\left(g(t)z(t)\right),\quad t\in J,
\end{equation}
and
\begin{equation*}
z(0)=0.
\end{equation*}
We have
\begin{eqnarray*}
z^{\Delta}(t)&=& L(t, x(t))\\ \\
&\leq& L\left(t, f(t)+\psi(g(t)z(t))\right)\\ \\
&=& L\left(t, f(t)+\psi(g(t) z(t))\right)-L(t, f(t))+L(t, f(t))\\ \\
&\leq& M(t, f(t))\psi^{-1}\left(f(t)+\psi(g(t))z(t)-f(t)\right)+L(t, f(t))\\ \\
&=& M(t, f(t))\psi^{-1}\left(\psi(g(t)z(t))\right)+h_1(t)\\ \\
&=& h_1(t)+M(t, f(t))g(t) z(t)\\ \\
&=& h_1(t)+h_2(t) z(t),\quad t\in J.
\end{eqnarray*}
Hence, we get
\begin{equation*}
z(t)\leq \int_a^t h_1(s) e_{\ominus h_2}(\sigma(s), t) \Delta s,\quad t\in J.
\end{equation*}
From here and from \eqref{5.5}, we obtain
\begin{equation*}
x(t)\leq f(t)+\psi\left(g(t)\int_a^t h_1(s) e_{\ominus h_2}(\sigma(s), t)\Delta s\right),\quad t\in J.
\end{equation*}
This completes the proof.

Lemma

Let \(p\geq q\geq 0\), \(p\ne 0\), and \(a\geq 0\). Then, for any \(k>0\), we have
\begin{equation*}
a^{q\over p}\leq \frac{q}{p} k^{\frac{q-p}{p}}a+\frac{p-q}{p}k^{q\over p}.
\end{equation*}

Proof

1. Let \(a=0\). Then
\begin{equation*}
0\leq \frac{p-q}{p} k^{q\over p}.
\end{equation*}

2. Let \(a>0\). Set
\begin{equation*}
f(k)= \frac{q}{p} k^{\frac{q-p}{p}}a+\frac{p-q}{p}k^{q\over p},\quad k>0.
\end{equation*}
Then
\[\begin{eqnarray*}
f^{\prime}(k)&=& \frac{q(q-p)}{p^2} k^{\frac{q-2p}{p}}a+\frac{p-q}{p^2} k^{\frac{q-p}{p}}\\ \\
&=& \frac{q(p-q)}{p^2} k^{\frac{q-2p}{p}}(k-a),\quad k>0.
\end{eqnarray*}\notag\]
Hence, we conclude that
\begin{equation*}
f(k)\geq f(a),\quad k>0,
\end{equation*}
and since
\[\begin{eqnarray*}
f(a)&=& \frac{q}{p}\left(a^{\frac{q-p}{p}}\right)a+\frac{p-q}{p} a^{q\over p}\\ \\
&=& \frac{q}{p} a^{q\over p} +\frac{p-q}{p} a^{q\over p}\\ \\
&=& a^{q\over p},
\end{eqnarray*}\notag\]
we get the desired result. This completes the proof.

Theorem

Let \(x, f, g, h\in \mathcal{C}_{rd}(J)\) be nonnegative  functions, \(f\) and \(g\) be nondecreasing functions, and \(p>1\). Then, the inequality
\[
(x(t))^p \leq f(t)+g(t)\int\limits_a^t h(s) x(s)\Delta s,\quad t\in J,
\notag\]
implies the inequality
\[
x(t)\leq \left(f(t)+g(t)H_1(t)+g(t)\int\limits_a^t H_1(s)H_2(s) e_{\ominus H_2}(\sigma(s), t) \Delta s\right)^{1\over p},\quad t\in J,
\notag\]
where
\[\begin{eqnarray*}
H_1(t)&=& \frac{p-1}{p} k^{1\over p} \int\limits_a^t h(s)\Delta s+\frac{1}{p} k^{\frac{1-p}{p}}\int\limits_a^t h(s) f(s) \Delta s,\\ \\
H_2(t)&=& \frac{1}{p} k^{\frac{1-p}{p}}h(t) g(t),\quad t\in J,
\end{eqnarray*}\notag\]
for any \(k>0\).

Proof

Let
\[
z(t)=\left(f(t)+g(t)\int\limits_a^t h(s) x(s)\Delta s\right)^{1\over p},\quad t\in J.
\notag\]
For any \(t_1, t_2\in J\), one has
\[\begin{eqnarray*}
z(t_1)&=& \left(f(t_1)+g(t_1)\int\limits_a^{t_1}h(s) x(s)\Delta s\right)^{1\over p}\\ \\
&\geq& \left(f(t_2)+g(t_2)\int\limits_a^{t_2} h(s) x(s)\Delta s\right)^{1\over p}\\ \\
&=& z(t_2).
\end{eqnarray*}\notag\]
Therefore, the function \(z\) is a nonnegative and nondecreasing function and
\[
x(t)\leq z(t),\quad t\in J.
\notag\]
Hence,
\[\begin{eqnarray*}
z(t)&=& \left(f(t)+g(t)\int\limits_a^t h(s) x(s)\Delta s\right)^{1\over p}\\ \\
&\leq& \left(f(t)+g(t)\int\limits_a^t h(s) z(s)\Delta s\right)^{1\over p},\quad t\in J.
\end{eqnarray*}\notag\]
Set
\[
u(t)=\int\limits_a^t h(s) z(s)\Delta s,\quad t\in J.
\notag\]
Then
\[
z(t)\leq \left(f(t)+g(t)u(t)\right)^{1\over p},\quad t\in J.
\notag\]
Applying the above for \(q=1\) and
\[
a(t)= f(t)+g(t)u(t),\quad t\in J,
\notag\]
we find
\[\begin{eqnarray*}
z(t)&\leq& (f(t)+g(t)u(t))^{1\over p}\\ \\
&\leq& \frac{p-1}{p} k^{1\over p}+\frac{1}{p} k^{\frac{1-p}{p}}(f(t)+g(t)u(t)),\quad t\in J,\quad \mbox{for any}\quad k>0.
\end{eqnarray*}\notag\]
Therefore
\[\begin{eqnarray*}
u(t)&=& \int\limits_a^t h(s) z(s)\Delta s\\ \\
&\leq& \int\limits_a^t h(s) \left(\frac{p-1}{p}k^{1\over p}+\frac{1}{p} k^{\frac{1-p}{p}}(f(s)+g(s)u(s))\right)\Delta s\\ \\
&=& \frac{p-1}{p}k^{1\over p}\int\limits_a^t h(s)\Delta s+\frac{1}{p} k^{\frac{1-p}{p}}\int\limits_a^t h(s) f(s)\Delta s\\ \\
&&+ \frac{1}{p} k^{\frac{1-p}{p}}\int\limits_a^t h(s) g(s) u(s)\Delta s\\ \\
&=& H_1(t)+\int\limits_a^t H_2(s) u(s)\Delta s,\quad t\in J,\quad \mbox{for any}\quad k>0.
\end{eqnarray*}\notag\]
Now, applying the Gronwall inequality, we arrive at
\[
u(t)\leq H_1(t) +\int\limits_a^t H_1(s) H_2(s) e_{\ominus H_2}(\sigma(s), t) \Delta s,\quad t\in J,\quad \mbox{for any} \quad k>0,
\notag\]
and
\[\begin{eqnarray*}
(z(t))^p&\leq& f(t)+g(t)u(t)\\ \\
&\leq& f(t)+g(t)H_1(t)+g(t) \int\limits_a^t H_1(s) H_2(s) e_{\ominus H_2}(\sigma(s), t)\Delta s,\quad t\in J,\quad \mbox{for any} \quad k>0.
\end{eqnarray*}\notag\]
Then
\[\begin{eqnarray*}
x(t)&\leq& z(t)\\ \\
&\leq& \left( f(t)+g(t)H_1(t)+g(t) \int\limits_a^t H_1(s) H_2(s) e_{\ominus H_2}(\sigma(s), t)\Delta s\right)^{1\over p},\quad t\in J,\quad \mbox{for any} \quad k>0.
\end{eqnarray*}\notag\]
This completes the proof.

Corollary

Let \(x, f, g, h\in \mathcal{C}_{rd}(J)\) be nonnegative  functions, \(f\) and \(g\) be nondecreasing functions, and \(p>1\). Then, the inequality
\[
(x(t))^p \leq f(t)+g(t)\int\limits_a^t h(s) x(s)\Delta s,\quad t\in J,
\notag\]
implies the inequality
\[
x(t)\leq \left(f(t)+g(t)H_1(t)e_{H_2}(t, a)\right)^{1\over p},\quad t\in J,
\notag\]
where
\[\begin{eqnarray*}
H_1(t)&=& \frac{p-1}{p} k^{1\over p} \int\limits_a^t h(s)\Delta s+\frac{1}{p} k^{\frac{1-p}{p}}\int\limits_a^t h(s) f(s) \Delta s,\\ \\
H_2(t)&=& \frac{1}{p} k^{\frac{1-p}{p}}h(t) g(t),\quad t\in J,
\end{eqnarray*}\notag\]
for any \(k>0\).

Proof.
We will use the notations in the proof of the last theorem. Observe that for any \(t_1, t_2\in J\), \(t_1\geq t_2\), one has
\[\begin{eqnarray*}
H_1(t_1)&=& \frac{p-1}{p}k^{1\over p}\int\limits_a^{t_1}h(s) \Delta s+\frac{1}{p}k^{\frac{1-p}{p}}\int\limits_a^{t_1} h(s) f(s)\Delta s\\ \\
&\geq& \frac{p-1}{p}k^{1\over p}\int\limits_a^{t_2}h(s) \Delta s+\frac{1}{p}k^{\frac{1-p}{p}}\int\limits_a^{t_2} h(s) f(s)\Delta s\\ \\
&=& H_1(t_2),\quad \mbox{for any}\quad k>0.
\end{eqnarray*}\notag\]
Thus, the function \(H_1\) is nondecreasing on the interval \(J\). Now, we arrive at
\[\begin{eqnarray*}
x(t)&\leq& \left(f(t)+g(t)H_1(t)+g(t)\int\limits_a^t H_1(s) H_2(s) e_{\ominus H_2}(\sigma(s), t) \Delta s\right)^{1\over p}\\ \\
&\leq& \left(f(t)+g(t)H_1(t)+g(t)H_1(t)\int\limits_a^t H_2(s) e_{\ominus H_2}(\sigma(s), t)\Delta s\right)^{1\over p}\\ \\
&=& \left(f(t)+g(t)H_1(t)+g(t)H_1(t) \int\limits_a^t \frac{H_2(s)}{1+\mu(s)H_2(s)}e_{\ominus H_2}(s, t)\Delta s\right)^{1\over p}\\ \\
&=& \left(f(t)+g(t) H_1(t)-g(t)H_1(t) \int\limits_a^t e_{\ominus H_2}^\Delta (s, t)\Delta s\right)^{1\over p}\\ \\
&=& \left(f(t)+g(t)H_1(t) -g(t)H_1(t) e_{\ominus H_2}(s, t)\bigg|_{s=a}^{s=t}\right)\\ \\
&=& \left(f(t)+g(t)H_1(t)-g(t)H_1(t)+g(t)H_1(t) e_{\ominus H_2}(s, t)\right)^{1\over p}\\ \\
&=& \left(f(t) +g(t)H_1(t) e_{H_2}(t, a)\right)^{1\over p},\quad t\in J,\quad \mbox{for any}\quad k>0.
\end{eqnarray*}\notag\]
This completes the proof.

Corollary

Let \(x, f, g, h\in \mathcal{C}_{rd}(J)\) be nonnegative  functions, \(f\) and \(g\) be nondecreasing functions, and \(q>1\). Then, the inequality
\[
x(t) \leq f(t)+g(t)\int\limits_a^t h(s) (x(s))^{1\over q}\Delta s,\quad t\in J,
\notag\]
implies the inequality
\[
x(t)\leq f(t)+g(t)H_1(t)+g(t)\int\limits_a^t H_1(s)H_2(s) e_{\ominus H_2}(\sigma(s), t) \Delta s,\quad t\in J,
\notag\]
where
\[\begin{eqnarray*}
H_1(t)&=& \frac{q-1}{q} k^{1\over q} \int\limits_a^t h(s)\Delta s+\frac{1}{q} k^{\frac{1-q}{q}}\int\limits_a^t h(s) f(s) \Delta s,\\ \\
H_2(t)&=& \frac{1}{q} k^{\frac{1-q}{q}}h(t) g(t),\quad t\in J,
\end{eqnarray*}\notag\]
for any \(k>0\).

Proof
Let
\[
y(t)=(x(t))^{1\over q},\quad t\in J.
\notag\]
Then
\[
x(t)=(y(t))^q,\quad t\in J,
\notag\]
and the given inequality takes the form
\[
(y(t))^q\leq f(t)+g(t) \int\limits_a^t h(s) y(s)\Delta s,\quad t\in J.
\notag\]
Then,  we get
\[
y(t)\leq \left(f(t)+g(t) H_1(t)+g(t)\int\limits_a^t H_1(s) H_2(s) e_{\ominus H_2}(\sigma(s), t) \Delta s\right)^{1\over q},\quad t\in J,
\notag\]
for any \(k>0\), or
\[
(x(t))^{1\over q}\leq \left(f(t)+g(t)H_1(t)+g(t) \int\limits_a^t H_1(s)H-2(s0 e_{\ominus H_2}(\sigma(s), t) \Delta s\right)^{1\over q},\quad t\in J,
\notag\]
for any \(k>0\), whereupon we get the desired result. This completes the proof.

Corollary

Let \(x, f, g, h\in \mathcal{C}_{rd}(J)\) be nonnegative  functions, \(f\) and \(g\) be nondecreasing functions, and \(q>1\). Then, the inequality
\[
(x(t))^p \leq f(t)+g(t)\int\limits_a^t h(s) x(s)\Delta s,\quad t\in J,
\notag\]
implies the inequality
\[
x(t)\leq f(t)+g(t) H_1(t) e_{H_2}(t, a),\quad t\in J,
\notag\]
where
\[\begin{eqnarray*}
H_1(t)&=& \frac{q-1}{q} k^{1\over q} \int\limits_a^t h(s)\Delta s+\frac{1}{q} k^{\frac{1-q}{q}}\int\limits_a^t h(s) f(s) \Delta s,\\ \\
H_2(t)&=& \frac{1}{q} k^{\frac{1-q}{q}}h(t) g(t),\quad t\in J,
\end{eqnarray*}\notag\]
for any \(k>0\).

Proof
We have
\[
x(t)\leq f(t)+g(t)H_1(t)+g(t)\int\limits_a^t H_1(s) H_2(s) e_{\ominus H_2}(\sigma(s), t)\Delta s,\quad t\in J,
\notag\]
for any \(k>0\), whereupon we get the desired result. This completes the proof.